Problem 20 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 1} \frac{x \ln x}{x^{2}-1} $$

Step-by-Step Solution

Answer

The limit is \(\frac{1}{2}\).

Step 1: Identify the Indeterminate Form

We first substitute the value of the limit into the expression. When we replace \(x = 1\), the expression becomes \(\frac{1 \cdot \ln(1)}{1^2 - 1} = \frac{0}{0}\), indicating an indeterminate form of type \(\frac{0}{0}\). This suggests that L'Hôpital's rule could be applied.

Step 2: Apply L'Hôpital's Rule

Since the limit leads to an indeterminate form \(\frac{0}{0}\), we apply L'Hôpital's Rule, which states that \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided the latter limit exists. Hence, we differentiate the numerator and the denominator separately.

Step 3: Differentiate the Numerator and Denominator

For the numerator \(f(x) = x \ln x\), using the product rule, we find \(f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\). For the denominator \(g(x) = x^2 - 1\), we have \(g'(x) = 2x\).

Step 4: Compute the Limit of the New Expression

Applying the derivatives obtained, the limit becomes \(\lim_{x \to 1} \frac{\ln x + 1}{2x}\). Substituting \(x = 1\) into this expression gives \(\frac{\ln(1) + 1}{2 \times 1} = \frac{0 + 1}{2} = \frac{1}{2}\).

Step 5: Verify the Solution

After calculating the limit using L'Hôpital's Rule, the result is \(\frac{1}{2}\), which is a determinate value confirming the limit evaluation was correct.

Key Concepts

Indeterminate Forms

In calculus, the concept of indeterminate forms is essential when evaluating limits. Indeterminate forms appear when substituting a limit value into a function results in expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(1^\infty\), etc. These forms are termed 'indeterminate' because they do not provide enough information to determine the actual limit value.

To resolve this ambiguity and find the limit, additional techniques like algebraic manipulation, series expansion, or rules like L'Hôpital's Rule are necessary. Recognizing an indeterminate form is the first crucial step in the process. For instance, in the exercise given, substituting \(x = 1\) into \(\frac{x \ln x}{x^{2}-1}\) yields \(\frac{0}{0}\). This indicates an indeterminate form is present, guiding the solver to seek further methods to compute this limit.

L'Hôpital's Rule

L'Hôpital's Rule offers a powerful and systematic way to resolve limits that present indeterminate forms. This rule can be applied to limits producing forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), transforming them into potentially simpler calculations.

The rule states that for functions \(f(x)\) and \(g(x)\), if \(\lim_{x \to c} f(x) / g(x)\) is an indeterminate form, then:

• \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\), assuming the latter limit exists.

In the problem example, after confirming the indeterminate form \(\frac{0}{0}\), the next step is to apply L'Hôpital’s Rule. Differentiating the numerator and the denominator results in a new limit expression \(\lim_{x \to 1} \frac{\ln(x) + 1}{2x}\), which is more straightforward to evaluate. This illustrates how L'Hôpital’s Rule turns a complex limit problem into a simpler one, enabling the solver to find the actual limit value.

Differentiation

Differentiation, a fundamental concept in calculus, involves finding the derivative of a function. The derivative measures how a function changes as its input changes, essentially providing the slope of the function at any given point.

In the context of solving limits using L'Hôpital's Rule, differentiation is employed to transform indeterminate forms into solvable expressions. The process requires differentiating both the numerator and the denominator separately. The derivative of the numerator in our example, \(x \ln x\), is found using the product rule: \(f'(x) = \ln x + 1\). For the denominator \(x^2 - 1\), the derivative is \(g'(x) = 2x\).

• The product rule is used when differentiating products of functions, as seen with the numerator \(x \ln x\).
• Simplification follows, leading to the possibility of easily evaluating limits.

Mastering differentiation, therefore, is key to applying L'Hôpital's Rule effectively, allowing complex limit problems to be tackled with confidence.